Ciucu–Krattenthaler enumeration conjecture for stationary-state coefficients

From papers

Write pp repeated opening parentheses as (p(^p and pp repeated opening parentheses followed by pp closing parentheses as ()p=(p)p(\ldots)_p=(^p\ldots)^p. For matchings of the form (_s_t)_p, write [s,t,p][s,t,p]. Let aFa_F be the coefficient of matching FF in the stationary state of H2p+2s+2tCH^{\rm C}_{2p+2s+2t}. Ciucu–Krattenthaler stationary-state coefficient conjecture. For F=[s,t,p]F=[s,t,p],

a[s,t,p]=det1i,js((2(s+t+p)+j2is+tj)(2(s+t+p)+j2is+tj2i+1))a_{[s,t,p]}=\det_{1\leq i,j\leq s}\left(\binom{2(s+t+p)+j-2i}{s+t-j}-\binom{2(s+t+p)+j-2i}{s+t-j-2i+1}\right)

and equivalently

a[s,t,p]=j=1s(j1)!(2t+2p+2j1)!(2p+2j)j(3t+2p+3j)sj(t+2p+s+2j1)!(t+sj)!.a_{[s,t,p]}=\prod_{j=1}^{s}\frac{(j-1)!(2t+2p+2j-1)!(2p+2j)_j(3t+2p+3j)_{s-j}}{(t+2p+s+2j-1)!(t+s-j)!}.

The claim is attributed to Mitra et al. and uses enumerations of hexagons with cut-off corners by Ciucu and Krattenthaler. Its resolution is not stated in the supplied text.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

Jan de Gier, “Loops, matchings and alternating-sign matrices”, arXiv:math/0211285 (2003).

Solutions 0

No solutions have been posted yet.